Answered: V (m/s) 12.0 8.00 4.00 t(s) 2.00 4.00…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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$### Apparent Weight Calculation in an Elevator #### Problem Statement: A 65.0 kg passenger is in an elevator. Using the velocity graph provided below, determine the passenger's apparent weight at $ t = 7.0 $ seconds. Assume the positive direction is upwards. #### Velocity Graph (V vs. t) The graph depicts the velocity $ V $ (in m/s) of the elevator as a function of time $ t $ (in seconds). The axes are labeled as follows: - $ V $ Axis (vertical): ranges from $ 0 $ to $ 12.0 $ m/s. - $ t $ Axis (horizontal): ranges from $ 0 $ to $ 12.0 $ seconds. The green line on the graph represents the velocity profile of the elevator: 1. From $ t = 0 $ to $ t = 8.0 $ seconds, the velocity increases linearly from $ 4.0 $ m/s to $ 12.0 $ m/s. 2. At $ t = 10 $ seconds, velocity drops sharply back to $ 0 $ m/s. #### Multiple-Choice Answers for Apparent Weight at $ t = 7.0 $: - $ 52 $ N - $ 585 $ N - $ 117 $ N - $ 689 $ N - $ 715 $ N #### Calculation Approach: 1. **Determine Acceleration:** - Between $ t = 0 $ and $ t = 8.0 $ seconds, the velocity increases from $ 4.0 $ m/s to $ 12.0 $ m/s. - Slope (acceleration, $ a $) of the velocity-time graph is calculated as: \[ a = \frac{\Delta V}{\Delta t} = \frac{12.0 \, \text{m/s} - 4.0 \, \text{m/s}}{8.0 \, \text{s} - 0 \, \text{s}} = 1.0 \, \text{m/s}^2 \] 2. **Calculate Apparent Weight at $ t = 7.0 $ seconds:** - Normal weight ($ W $) of the passenger is given by:$

Transcribed Image Text:### Apparent Weight Calculation in an Elevator #### Problem Statement: A 65.0 kg passenger is in an elevator. Using the velocity graph provided below, determine the passenger's apparent weight at $ t = 7.0 $ seconds. Assume the positive direction is upwards. #### Velocity Graph (V vs. t) The graph depicts the velocity $ V $ (in m/s) of the elevator as a function of time $ t $ (in seconds). The axes are labeled as follows: - $ V $ Axis (vertical): ranges from $ 0 $ to $ 12.0 $ m/s. - $ t $ Axis (horizontal): ranges from $ 0 $ to $ 12.0 $ seconds. The green line on the graph represents the velocity profile of the elevator: 1. From $ t = 0 $ to $ t = 8.0 $ seconds, the velocity increases linearly from $ 4.0 $ m/s to $ 12.0 $ m/s. 2. At $ t = 10 $ seconds, velocity drops sharply back to $ 0 $ m/s. #### Multiple-Choice Answers for Apparent Weight at $ t = 7.0 $: - $ 52 $ N - $ 585 $ N - $ 117 $ N - $ 689 $ N - $ 715 $ N #### Calculation Approach: 1. **Determine Acceleration:** - Between $ t = 0 $ and $ t = 8.0 $ seconds, the velocity increases from $ 4.0 $ m/s to $ 12.0 $ m/s. - Slope (acceleration, $ a $) of the velocity-time graph is calculated as: \[ a = \frac{\Delta V}{\Delta t} = \frac{12.0 \, \text{m/s} - 4.0 \, \text{m/s}}{8.0 \, \text{s} - 0 \, \text{s}} = 1.0 \, \text{m/s}^2 \] 2. **Calculate Apparent Weight at $ t = 7.0 $ seconds:** - Normal weight ($ W $) of the passenger is given by:

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